Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{-10r + 10}{-2r - 6} \div \dfrac{r^2 - 4r + 3}{-r + 3} $
Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-10r + 10}{-2r - 6} \times \dfrac{-r + 3}{r^2 - 4r + 3} $ First factor the quadratic. $n = \dfrac{-10r + 10}{-2r - 6} \times \dfrac{-r + 3}{(r - 1)(r - 3)} $ Then factor out any other terms. $n = \dfrac{-10(r - 1)}{-2(r + 3)} \times \dfrac{-(r - 3)}{(r - 1)(r - 3)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -10(r - 1) \times -(r - 3) } { -2(r + 3) \times (r - 1)(r - 3) } $ $n = \dfrac{ 10(r - 1)(r - 3)}{ -2(r + 3)(r - 1)(r - 3)} $ Notice that $(r - 3)$ and $(r - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 10\cancel{(r - 1)}(r - 3)}{ -2(r + 3)\cancel{(r - 1)}(r - 3)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $n = \dfrac{ 10\cancel{(r - 1)}\cancel{(r - 3)}}{ -2(r + 3)\cancel{(r - 1)}\cancel{(r - 3)}} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $n = \dfrac{10}{-2(r + 3)} $ $n = \dfrac{-5}{r + 3} ; \space r \neq 1 ; \space r \neq 3 $